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Let $z \neq-i$ be any complex number such that $\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}$ is a purely imaginary number.
Then $\mathrm{z}+\frac{1}{\mathrm{z}}$ is:
Options:
Then $\mathrm{z}+\frac{1}{\mathrm{z}}$ is:
Solution:
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Verified Answer
The correct answer is:
any non-zero real number.
any non-zero real number.
Let $z=x+i y$
$\frac{z-i}{z+i}$ is purely imaginary means its real part is zero.
$$
\begin{aligned}
&\frac{x+i y-i}{x+i y+i}=\frac{x+i(y-1)}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)} \\
&=\frac{x^2-2 i x(y+1)+x i(y-1)+y^2-1}{x^2+(y+1)^2} \\
&=\frac{x^2+y^2-1}{x^2+(y+1)^2}-\frac{2 x i}{x^2+(y+1)^2}
\end{aligned}
$$
for pure imaginary, we have
$$
\begin{aligned}
& \frac{x^2+y^2-1}{x^2+(y+1)^2}=0 \\
\Rightarrow & x^2+y^2=1 \\
\Rightarrow &(x+i y)(x-i y)=1 \\
\Rightarrow & x+i y=\frac{1}{x-i y}=z
\end{aligned}
$$
and $\frac{1}{z}=x-i y$
$$
z+\frac{1}{z}=(x+i y)+(x-i y)=2 x
$$
$\left(z+\frac{1}{z}\right)$ is any non-zero real number
$\frac{z-i}{z+i}$ is purely imaginary means its real part is zero.
$$
\begin{aligned}
&\frac{x+i y-i}{x+i y+i}=\frac{x+i(y-1)}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)} \\
&=\frac{x^2-2 i x(y+1)+x i(y-1)+y^2-1}{x^2+(y+1)^2} \\
&=\frac{x^2+y^2-1}{x^2+(y+1)^2}-\frac{2 x i}{x^2+(y+1)^2}
\end{aligned}
$$
for pure imaginary, we have
$$
\begin{aligned}
& \frac{x^2+y^2-1}{x^2+(y+1)^2}=0 \\
\Rightarrow & x^2+y^2=1 \\
\Rightarrow &(x+i y)(x-i y)=1 \\
\Rightarrow & x+i y=\frac{1}{x-i y}=z
\end{aligned}
$$
and $\frac{1}{z}=x-i y$
$$
z+\frac{1}{z}=(x+i y)+(x-i y)=2 x
$$
$\left(z+\frac{1}{z}\right)$ is any non-zero real number
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